Question

A mass of 4 kg is compressed 0.79 meters into a spring with a spring constant of 400 N/m and held still on an initially frictionless horizontal surface (see sketch). In front of the spring is a portion of the surface with friction which is 3.2 meters long and the coefficient of kinetic friction is 0.53. The spring is at the the top of a frictionless incline with a length of 26 meters and angle of 23 degrees. At the bottom of the incline is a frictionless horizontal surface and a frictionless circular loop of radius 5 meters. The mass travels on the inside of this loop. What is the amount of normal force acting on the mass on the side of the loop (halfway between the top and bottom) in Newtons?

Answer 1

height of incline h1 = L*sintheta = 26*sin23

final height h2 = R ( radius of loop)

work done by gravity Wg = m*g*(h1-h2)

work done by friction Wf = -uk*m*g*d

work done by elastic force in spring We = (1/2)*k*x^2

initial kinetic energy K1 = 0

final kinetic energy at half way K2 = (1/2)*m*v^2

from work energy relation

net work = change in kE

Wg + Wf + We = K2 - K1

m*g*(h1-h2) - uk*m*g*d + (1/2)*k*x^2 = (1/2)*m*v^2

4*9.8*(26*sin23 - 5) - (0.53*4*9.8*3.2) + (1/2)*400*0.79^2 = (1/2)*4*v^2

speed at half way v = 11.4 m/s

normal force Fn = m*v^2/R = 4*11.4^2/5 = 104 N <<<-------answer